Evaluate the definite integral: int_{frac{1}{ | vedclass

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(D) Let $I = \int_{\frac{1}{2}}^{2} \frac{x^2 \ln x}{(1+x^2)^3} dx$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we note that

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(D) Let $I = \int_{\frac{1}{2}}^{2} \frac{x^2 \ln x}{(1+x^2)^3} dx$.Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we note that the limits are reciprocal ($a=1/2, b=2$,so $a 	imes b = 1$).Let $x = \frac{1}{t}$,then $dx = -\frac{1}{t^2} dt$.When $x = \frac{1}{2}, t = 2$. When $x = 2, t = \frac{1}{2}$.Substituting these into the integral:$I = \int_{2}^{\frac{1}{2}} \frac{(\frac{1}{t})^2 \ln(\frac{1}{t})}{(1+(\frac{1}{t})^2)^3} (-\frac{1}{t^2}) dt$$I = \int_{\frac{1}{2}}^{2} \frac{\frac{1}{t^2} (-\ln t)}{(\frac{t^2+1}{t^2})^3} \frac{1}{t^2} dt$$I = \int_{\frac{1}{2}}^{2} \frac{-\ln t \cdot \frac{1}{t^4}}{\frac{(1+t^2)^3}{t^6}} dt$$I = \int_{\frac{1}{2}}^{2} \frac{-\ln t \cdot t^2}{(1+t^2)^3} dt$$I = -\int_{\frac{1}{2}}^{2} \frac{t^2 \ln t}{(1+t^2)^3} dt = -I$.Therefore,$2I = 0$,which implies $I = 0$.

Evaluate the definite integral: $\int_{\frac{1}{2}}^{2} \frac{x^2 \ln x}{(1+x^2)^3} dx$

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